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telo118 [61]
2 years ago
13

What is the mass of silver in 4.3g AgNO3?

Chemistry
1 answer:
gregori [183]2 years ago
5 0
4.22 grams.
1. First find out how much AgNO3 weighs with one mole (107.87 g Ag + 14.007 g N + 48 g O = 169.89 grams)
2. Find the percent of Ag you have. So, (107.87 g/mol Ag)/(169.89 g/mol AgNO3)= 0.63 * 100 = 63%.
3. If you have 6.7 grams total, you know 63% of it is going to be silver, so just multiply 6.7 grams by .63 and you get 4.22 g Ag
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Type the correct answer in the box. Express the answer to two significant figures.
meriva

Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol

8 0
3 years ago
Read 2 more answers
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
How do you do this?
Alona [7]
It would be Atom 2 because the proton and neutron are both nine and the election is 8 which is -1 to nine <span />
7 0
2 years ago
Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el
Alexxandr [17]

Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

Ni^{2+}+2e^-\rightarrow Ni

96500\times 2=193000Coloumb of electricity deposits 1 mole of Nickel

1 mole of Nickel weighs = 58.7 g

Given quantity = 18.0 kg = 18000 g  (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

18000 g of Nickel is deposited by = \frac{193000}{58.7}\times 18000=59182282.8C of electricity

Q=I\times t

where Q= quantity of electricity in coloumbs  = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

59182282.8C=31.0A\times t

t=1909105.9sec

(1h=3600 sec)

t=530hours

Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A

3 0
2 years ago
HELP HELP ME PLSS
Akimi4 [234]

Answer:

the average answer in maybe 60

6 0
2 years ago
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